LeetCode_Construct Binary Tree from Preorder and Inorder Traversal

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

　　

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
  TreeNode * constructSubtree(vector<int> &preOrder, int preStart, int preEnd, vector<int> &inOrder, int inStart, int inEnd){
      
    // if(preStart < preEnd) return NULL ;
       
     int temp = preOrder[preStart] ;
     TreeNode *root = new TreeNode(temp) ;
     if(preStart == preEnd)
        return root ;
        
     int i;
     for(i = inStart ; i<= inEnd ;i++)
       if(inOrder[i] == temp ) break;
      
     root->left = preStart+1 <= preStart+i-inStart ? constructSubtree(preOrder, preStart+1, preStart+i-inStart, inOrder, inStart, i-1) : NULL;
     root->right = preStart+i-inStart+1 <=  preEnd ? constructSubtree(preOrder, preStart+i-inStart+1,  preEnd, inOrder, i+1, inEnd) : NULL;
     
     return root ;
 } 
    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
    
        if(preorder.size() == 0 ) return NULL;
        TreeNode *root = new TreeNode(preorder[0]) ;
        if(preorder.size() == 1 ) return root ;
        
        int i;
        int temp = preorder[0] ;
        
        for(i = 0; i< inorder.size(); i++)
           if(inorder[i] == temp) break ;
        
         root->left = 1 <= i ? constructSubtree(preorder, 1,i,inorder, 0, i-1) : NULL  ;
    
         root->right = i+1 <= preorder.size()-1 ? constructSubtree(preorder, i+1,preorder.size()-1, inorder, i+1, inorder.size() -1): NULL ;
        
        return root;
   }
   
};